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1. In how many way could 15 different books be divided equally among 3 people?
Answer in the book = 756756

2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
Answer in the book = 459

3. There are 12 questions on an examination , and each student must answer 8 questions including at least 4 of the first 5 questions. How many different combinations of questions could a student choose to answer?
Answer in the book = 210
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1. In how many way could 15 different books be divided equally among 3 people?
Answer in the book = 756756

15 books can be divided 15! ways = 1307674368000

Even though the books are different, the order within the group is not important. ie. ABCDE = ADECB (both groups contain the same letters). Each group of 5 books can be divided in 5! ways.

Hence, 15 different books are divided eually among 3 people

ways.

2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
Answer in the book = 459

If you know what a venn diagram is, it might be helpful to draw problems like this using one...

There are 12 members all together (5 + 8 - 1(the common member) = 12)

These 12 members can be arranged in ways.

This number represents the total permutations for the group. Taking out the other groups which contain only members from one group:

If a student does all of the first 5 problems, then he must find 3 more problems to do out of the 7 remaining problems available. There are ways to do this.

If a student does only 4 of the first five, There are ways to select those four problems (think of it as choosing a problem to skip). After those 4, he must find 4 more problems to do out of the 7 remaining problems available. There are ways to do this. Multiply the ways to select the first four problems by the ways to select the last four problems to get .